走迷宮值路徑記錄

利用bfs走迷宮是最基礎的方法，記錄訪問過的每個點然後回溯最短路徑的點

首先把問題簡單化一下，只有牆壁和路，代碼比較簡單，只要會bfs基本上看得懂

// 走迷宮記錄路徑
// 概述：有一個N*M大小的迷宮，其中'#'代表牆壁無法通過，'.'代表路，每走一步都將花費1s的時間，求從起點(0,0)到終點(N-1,M-1)的最短路徑，並將路徑打印
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int INF = 1000005;
const int MAX_N = 105;
struct Point
{
    int x;
    int y;
}first, next,pre[MAX_N][MAX_N];
char pic[MAX_N][MAX_N];
int N, M;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int value[MAX_N][MAX_N];
bool vis[MAX_N][MAX_N];

int bfs()
{
    queue<Point> que;
    que.push(first);
    while(que.size())
    {
        first = que.front();
        que.pop();
        if(first.x == N - 1 && first.y == M -1)
            return value[N - 1][M - 1];
        for(int i = 0; i < 4; i++)
        {
            next.x = first.x + dx[i];
            next.y = first.y + dy[i];
            if(next.x >= 0 && next.x < N && next.y >= 0 && next.y < M && pic[next.x][next.y] != '#' && vis[next.x][next.y])
            {
                value[next.x][next.y] = value[first.x][first.y] + 1;
                pre[next.x][next.y].x = first.x;
                pre[next.x][next.y].y = first.y;
                vis[next.x][next.y] = false;
                que.push(next);
            }
        }
    }

    return INF;
}

void print_pre(int i, int j)
{
    if(pre[i][j].x == -1)
        return;
    print_pre(pre[i][j].x, pre[i][j].y);
    printf("%ds:(%d,%d)->(%d,%d)\n",value[i][j], pre[i][j].x, pre[i][j].y, i, j);
}

int main()
{
    while(cin >> N >> M)
    {
        for(int i = 0; i < N; i++)
            cin >> pic[i];
        first.x = 0;
        first.y = 0;
        memset(vis, true, sizeof(vis));     // true表示該點未訪問，false表示該點已訪問
        vis[0][0] = false;
        pre[0][0].x = pre[0][0].y = -1;
        int ans = bfs();
        if(ans == INF)
            printf("sorry,can't out!\n");
        else
        {
            printf("It takes %d seconds to reach the target position, let me show you the way.\n", ans);
            print_pre(N - 1, M - 1);
        }
    }

    return 0;
}

hangd1026題就是這種類型的題目，只是多了一個條件，路上可能會有怪物，每個怪物有N點生命值（1<=n<=9），要花費n秒殺死這個怪物。

題目鏈接

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int INF = 100005;
const int MAX_N = 105;
struct Node
{
    int x;
    int y;
    int num;
    bool operator<(const Node &a) const
    {
        return num > a.num;
    }
}first, nextNode;
struct Point
{
    int x;
    int y;
}pre[MAX_N][MAX_N];
char pic[MAX_N][MAX_N];
int value[MAX_N][MAX_N];
bool vis[MAX_N][MAX_N];
int N, M;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};

int bfs()
{
    priority_queue<Node> pque;
    pque.push(first);
    while(pque.size())
    {
        first = pque.top();
        pque.pop();
        if(first.x == N - 1 && first.y == M - 1)
            return value[N - 1][M - 1];
        for(int i = 0; i < 4; i++)
        {
            nextNode.x = first.x + dx[i];
            nextNode.y = first.y + dy[i];
            if(nextNode.x >= 0 && nextNode.x < N && nextNode.y >= 0 && nextNode.y < M && pic[nextNode.x][nextNode.y] != 'X' && vis[nextNode.x][nextNode.y])
            {
                if(pic[nextNode.x][nextNode.y] == '.')
                {
                    nextNode.num = first.num + 1;
                }
                else if(isdigit(pic[nextNode.x][nextNode.y]))
                {
                    nextNode.num = first.num + 1 + (int)(pic[nextNode.x][nextNode.y] - '0');
                }
                pre[nextNode.x][nextNode.y].x = first.x;
                pre[nextNode.x][nextNode.y].y = first.y;
                value[nextNode.x][nextNode.y] = nextNode.num;
                vis[nextNode.x][nextNode.y] = false;
                pque.push(nextNode);
            }
        }

    }

    return INF;
}

void print_pre(int i, int j)
{
    if(pre[i][j].x == -1)
        return;
    print_pre(pre[i][j].x, pre[i][j].y);
    if(pic[i][j] == '.')
        printf("%ds:(%d,%d)->(%d,%d)\n",value[i][j], pre[i][j].x, pre[i][j].y, i, j);
    else
    {

        int t = pic[i][j] - '0';
        int kase = value[pre[i][j].x][pre[i][j].y];
        printf("%ds:(%d,%d)->(%d,%d)\n",++kase, pre[i][j].x, pre[i][j].y, i, j);
        for(int k = 0; k < t; k++)
        {
            printf("%ds:FIGHT AT (%d,%d)\n", ++kase, i, j);
        }

    }
}

int main()
{
    while(cin >> N >> M)
    {
        for(int i = 0; i < N; i++)
            cin >> pic[i];
        memset(vis, true, sizeof(vis));
        first.x = 0;
        first.y = 0;
        first.num = 0;
        value[0][0] = 0;
        vis[0][0] = false;
        pre[0][0].x = pre[0][0].y = -1;
        int ans = bfs();
        if(ans != INF)
        {
            printf("It takes %d seconds to reach the target position, let me show you the way.\n", ans);
            print_pre(N - 1, M - 1);
        }
        else
            printf("God please help our poor hero.\n");
        printf("FINISH\n");
    }

    return 0;
}